|((x+1))/(x-2)|>2

Lösung

\frac{( x + 1 )}{x - 2} > 2

1 < x < 2 or 2 < x < 5

Intervall-Notation

(1, 2) \cup (2, 5)

Schritte zur Lösung

\frac{x + 1}{x - 2} > 2

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

\frac{x + 1}{x - 2} < - 2 or \frac{x + 1}{x - 2} > 2

\frac{x + 1}{x - 2} < - 2 or \frac{x + 1}{x - 2} > 2

1 < x < 2 or 2 < x < 5

x^{2} + 14 x + 1 = 0

3 x^{2} + 8 x - 16 = 0

x + 2 < 2 x - 3 \leq x + 1

s im pl i f y e^{- \frac{1}{4}}

\frac{x}{[ - 4 ]} = - 12